# Tag Info

12

You'll need to understand the sampling theorem. In short, each signal has what we call a spectrum¹, which is the Fourier transform of the signal as it comes in time domain (if it is a time signal), or spatial domain (if it is a picture. Since the Fourier transform is bijective, a signal and its transform are equivalent; in fact, one can often interpret the ...

11

Suppose that the sampling is done at a rate of $1000$ Hz, one sample every millisecond. Suppose also that the signal being sampled is at $3200$ Hz, the first sample is at the peak of the sinusoid. The next sample, will be taken one millisecond later during which time the sinusoid will have gone through $3.2$ periods, and so the next sample will have the ...

8

One word for that technique is superresolution. Robert Gawron has a blog post here and the Python implementation here. Usually, this technique relies on each image being slightly offset from the others. The only gain you'd get from not moving between shots would be to reduce the noise level.

8

You can't identify aliasing with a simple sinusoid at a specific frequency and that in a way is the whole point about trying to avoid it. You can't know if the sinusoid you are "looking at" is $Q$ Hz or $2Fs-Q$Hz. A single aliased sinusoidal component looks just like a non-aliased sinusoid. If you want to experience aliasing, you have to attempt it with ...

7

The reason is that if it is true for any $m$, it is also true for $m=kn$. I will sketch the proof in another way. Call $f_s = 1/t_s$ sampling frequency where $t_s$ is sampling period, the two signal $x(t) = \sin(2\pi f_0 t)$ and $x_k(t) = \sin(2\pi (f_0 + k f_s) t)$ have the same values at sampling instants (aliasing), i.e. $x[n] = x_k[n]$. Indeed, \...

6

Regarding example 1: first of all, either the fft or the ifft needs to be normalised by the number of sampling points as you have done (actually you can normalise each by a factor of the square root of the number of points, it is just a meter of definition). However, in your case the ifft is half the length of the fft you performed. Hence your signal is ...

6

You have the right idea, but a key mistake was messing up your results. n=0:1/Fs:1; By making the time increments 1/Fs you are implicitly making the time units seconds. In other words, "0" is 0 seconds, and "1" is 1 second. Cont = A*sin(2*pi*(Fmax/Fs)*n); By normalizing the sinusoid frequency by Fs you are implicitly assuming that the sample unit is 1/...

6

I think you mean "images", not "aliases". They become aliases if there is foldover from resampling. It's because you are not adding two signals, $x(t)$ and $\operatorname{III}(t)$, you are multiplying them that these images appear. \begin{align} x_\text{s}(t) & \triangleq x(t) \cdot \operatorname{III}(t/T) \\ &= x(t) \cdot \sum\limits_{n=-\... 5 Applying a non-linear function will always introduce harmonics, and mixing non-linear functions with sampled versions of continuous signals does add the wrinkle you note above (where high-frequency harmonics are aliased to low frequencies.) I can think of a few ways to proceed: You can use an oversampling factor high enough to capture the extra harmonics (... 5 The concept of reconstruction has nothing to do with the application, rather it has to do with the question: did I get the same signal that is really there. If you cannot recreate the signal back, that means the conversion process is loosing/modifying underlying information, which in most cases you do not want to happen. So the confidence on the Fourier ... 5 You may not need to explicitly reconstruct. But if you did reconstruct a waveform using the samples that you have, and end up with something different from the actual input, your controller is controlling as if that new different reconstructed waveform was really the input. Depending on what your controller is doing, you may have wanted it to do something ... 5 The answer is: yes, sampling in the frequency domain causes aliasing in the time domain, exactly like the dual case: sampling in the time domain causes aliasing in the frequency domain. There are many ways to see this. One standard way is to sample the discrete-time Fourier transform (DTFT) of a discrete-time signal by multiplying it with a Dirac comb and ... 5 In addition to @hotpaw2 explanation, a graphic. There are two analog square waves (red and green), with different lengths. They are depicted with a fine sampling, denoted by crosses. Their actual sampling is denoted by circles. The red one is shorter than the green one, as can be seen in the interval ]0.7\;0.8[. Yet, the sample points are the same. Thus, ... 5 this paper was done long before i had MATLAB. the drawings are poor, but the math (at least in this revision, which is what you should use) is spot on. send me an email address (to my audioimagination.com, not the wavemechanics.com on the paper) and i will send you a very short C file that shows how to generate and crossfade the wavetables in the synthesis ... 4 A few approaches to alias-free nonlinear distortion (in increasing order of difficulty): Subband distortion: Use a low pass filter to extract the lower end of the signal. If you choose a cutoff frequency of \frac{f_s}{2N} you can apply any non-linear transfer function f with derivatives starting at f^{N+1} vanishing to avoid aliasing. Add just the ... 4 Intuitively, if You move the sensor  N  steps each at the size of  \frac{1}{N}  of its resolution you can get  \times N  more resolution. It is like a polyphase representation of the signal. Using estimation methods, any movement which is not an (Event with zero probability) integer multiplication of the resolution of the sensor, namely, fractional ... 4 Let’s look at an example. I’m going to use a sample rate of 50 throughout this post (i.e. the functions are sampled at 0, 0.02, 0.04, 0.06, …) and simple sine waves of the form f_a(t) = \sin (2\pi\cdot a\cdot t), where a is the frequency of the sine wave. For example, this is a plot of f_3: The red impulses indicate the sample values. The green line, ... 4 Think of frequency in this context more in terms of angular velocity. If that velocity is constant, nothing changes from the usual picture. However, if the frequency changes, it is more correct to compute the actual angle that is input into the trig functions as an anti-derivative of the frequency function. Here the interpolated frequency isf(t)=(1-\...

4

I think you should plot something like: t = [0:0.05:1]; %20Hz sampling a = sin(2*pi*2*t); %2Hz sine wave b = sin(2*pi*18*t); %18Hz sine wave plot(t, a, 'bo'); hold on; plot(t, b, 'ro'); T = [0:0.001:1]; %1000Hz sampling frequency A = sin(2*pi*2*T); B = sin(2*pi*18*T); % plot for 1000Hz sampling frequency plot(T, A, 'b'); plot(T, B, 'r'); ...

4

Looking at the spectrogram, the prominent artifacts go up or down in frequency synchronously to the 138 MHz signal but have larger bandwidths. That is an indication that they are its harmonics, due to a nonlinearity somewhere in the system. With the sampling frequency of 500 MHz some of the positive and negative harmonics alias to the following (bold) ...

4

This question, and a number of similar others, from DTSP book can be a little tricky to recognise the fact that it's not actually asking alias free operation (which would require 8 khz cutoff signal as you expected), rather it is actually asking how much aliasing (due to initial sampling block) is tolerable: you can allow aliasing in those regions of the DT ...

4

I read Footnote 8, "The complication is that because of the sampling, the total system is not time-translation invariant and so does not have a unique ‘impulse response’ – the response is slightly different according to the position of an original impulse relative to the sampling points." as a cumbersome way of saying that aliasing corrputs an LTI ...

4

This may not fully answer your question, but for getting a feeling of aliasing, maybe a simple demonstration can be helpful... Some initial setup: import numpy as np from matplotlib import pyplot as plt sr = 44100 sig_len = 4096 So, it is well known, that time domain undersampling causes frequency domain aliasing. # sine wav and its spectrum t = np....

4

Hints: Can an LTI system generate components in some frequency $\omega_0$ if the input signal $x(n)$ was such that $X(e^{j\omega_0})=0$? Does aliasing do such thing? The answers to these questions are straightforward and, combined, they answer the original question.

4

The name of this effect is Spectral Leakage. Remember the relation $$\frac{k}{N} = \frac{F}{F_S}$$ where $k$ is the bin number, $N$ is the FFT size, $F$ is the continuous frequency in Hz and $F_S$ is the sample rate in Hz. It can be seen that $k$ varies from $-N/2$ to $N/2-1$ (or from $0$ to $N-1$). So there are only $N$ continuous frequencies $F$ for which ...

4

When discussing DFT, you have to remember two things: you're windowing your true signal $x[n]$ (which is periodic, and then infinitely long) with a window $w[n]$ (here a rectangular window) to obtain a truncated version $x'[n]$ of it $$x'[n] = x[n]w[n].$$ you're sampling the Fourier Transform $X'(e^{j\Omega})$ of your signal at normalized pulsation $\... 4 Decimating a signal (selecting every Dth sample and discarding the rest) does not distort the signal within the passband in any way other than to cause aliases from higher frequencies to fold into the signal bandwidth. Depending on how we model the system the phase may be effected since$z^{-n}$is replaced with$z^{-n/D}$, but the phase will still be ... 4 This visual phenomenon appears because the maximum frequency is close to the Nyquist frequency, or half the sampling frequency. Sampling begins to approach the limit of$2$samples per period, and thus the linear interpolation performed by Matlab becomes highly inaccurate. However, samples are correctly located, as you can see from the code where an higher ... 3 Yes, there is something to it. The phase increment of each sample of the 2 Hz wave is$2\pi * 2*.05=.2\pi$. The phase increment of each sample of the 18 Hz wave is$2\pi*18*.05=1.8\pi=-.2\pi$. Thus, the 2 Hz wave is$sin(2\pi*2*t)$, and the 18 Hz wave is$sin(2\pi*-2*t)=-sin(2\pi*2*t)\$. Thus, the two waves are 180 degrees out of phase.

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